Amazing But Quite Confusing Series

Similar solution with Dhanvanth Balakrishnan 's

Let $\displaystyle S=1+\frac {2^3}{1!} +\frac {3^3}{2!} +\frac {4^3}{3!}+\dots = \sum_{n=0}^\infty \frac {(n+1)^3}{n!}$ . Consider the following:

$\begin{aligned} \sum_{n=0}^\infty \frac {x^n}{n!} & = e^x & \small \color{#3D99F6} \text{Multiply both sides by }x \\ \sum_{n=0}^\infty \frac {x^{n+1}}{n!} & = xe^x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \sum_{n=0}^\infty \frac {(n+1)x^{n}}{n!} & = e^x + xe^x & \small \color{#3D99F6} \text{Multiply both sides by }x \\ \sum_{n=0}^\infty \frac {(n+1)x^{n+1}}{n!} & = xe^x + x^2e^x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \sum_{n=0}^\infty \frac {(n+1)^2x^{n}}{n!} & = e^x + 3xe^x + x^2e^x & \small \color{#3D99F6} \text{Multiply both sides by }x \\ \sum_{n=0}^\infty \frac {(n+1)^2x^{n+1}}{n!} & = xe^x + 3x^2e^x + x^3e^x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \sum_{n=0}^\infty \frac {(n+1)^3x^{n}}{n!} & = e^x + 6xe^x + 7x^2e^x + x^3e^x & \small \color{#3D99F6} \text{Put }x=1 \\ \sum_{n=0}^\infty \frac {(n+1)^3}{n!} & = 15e \approx \boxed{40.774} \end{aligned}$

We know that the expansion of exponential function is,

$\large e^{x}$ = $\large 1$ $\large +$ $\large x$ $\large +$ $\large \frac{x^{2}}{2!}$ $\large +$ $\large \frac{x^{3}}{3!}$ $\large +$ $\large \frac{x^{4}}{4!}$ $\large +$ $\large ...$

By multiplying $\large x$ on both sides,

$\large xe^{x}$ = $\large x$ $\large +$ $\large x^{2}$ $\large +$ $\large \frac{x^{3}}{2!}$ $\large +$ $\large \frac{x^{4}}{3!}$ $\large +$ $\large \frac{x^{5}}{4!}$ $\large +$ $\large ...$

Differentiating w.r.t $\large x$ on both sides,

$\large xe^{x}+e^{x}$ = $\large 1$ $\large +$ $\large 2x$ $\large +$ $\large \frac{3x^{2}}{2!}$ $\large +$ $\large \frac{4x^{3}}{3!}$ $\large +$ $\large \frac{5x^{4}}{4!}$ $\large +$ $\large ...$

So by repeating this process $\large 2$ times, we get,

$\large 7e^{x}$ $\large +$ $\large 6x^{2}e^{x}$ $\large +$ $\large e^{x}$ $\large +$ $\large x^{3}e^{x}$ $\large =$ $\large 1$ $\large +$ $\large 2^{3}x$ $\large +$ $\large \frac{3^{3}x^{2}}{2!}$ $\large +$ $\large \frac{4^{3}x^{3}}{3!}$ $\large +$ $\large \frac{5^{3}x^{4}}{4!}$ $\large +$ $\large ...$

By substituting $\large x$ $\large =$ $\large 1$ ,

$\large 1$ $\large +$ $\large \frac{2^{3}}{1!}$ $\large +$ $\large \frac{3^{3}}{2!}$ $\large +$ $\large \frac{4^{3}}{3!}$ $\large +$ $\large ...$ $\large =$ $\large 15e$ $\large =$ $\large 40.774$

$\displaystyle I=1+\frac {2^3}{1!} +\frac {3^3}{2!} +\frac {4^3}{3!} +......= \sum_{n=0}^\infty \frac {(n+1)^3}{n!}$

Now, let $(n+1)^3=a [n(n-1)(n-2)]+b [n(n-1)]+c n+d$

$\implies n^3+3 n^2+3 n+1= a n^3+(b-3 a) n^2+(2 a-b+c)n+d$

From this identity, one can easily get, $a=1, b=6, c=7, d=1$ .

Then, $\begin{aligned} I&=1+\frac {2^3}{1!} +\frac {3^3}{2!}+\sum_{n=3}^\infty \frac {(n+1)^3}{n!}&\\ &=1+\frac {2^3}{1!} +\frac {3^3}{2!}+\sum_{n=3}^\infty \Big[\frac {a}{(n-3)!}+\frac {b}{(n-2)!}+\frac {c}{(n-1)!}+\frac {d}{n!}\Big]&\\ &=1+\frac {2^3}{1!} +\frac {3^3}{2!}+a\sum_{m=0}^\infty \frac {1}{m!}+b\Big(\sum_{m=0}^\infty \frac {1}{m!}-1\Big)+c\Big(\sum_{m=0}^\infty \frac {1}{m!}-1-1\Big)+d\Big(\sum_{m=0}^\infty \frac {1}{m!}-1-1-\frac{1}{2!}\Big)&\\ &=\Big(9-b-2c-2d+\frac{27-d}{2}\Big)+(a+b+c+d)e\quad\small \color{#3D99F6} \Big[\text{as } e=\sum_{m=0}^\infty \frac {1}{m!}\Big]&\\ \end{aligned}$

Now, putting the values of $a,b,c,d$ we get, $\boxed{I=15e=40.774}$